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\usepackage{amsmath}%数学方程的显示\usepackage{listings}%插入代码
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\lstset{
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\geometry{a4paper,left=2cm,right=2cm,top=2cm,bottom=2cm}%一定要放在前面！
\pagestyle{fancy}%设置页眉页脚
\lhead{陈冠宇\ 3200102033}%页眉左
\chead{cse}%页眉中
\rhead{}%章节信息
\cfoot{\thepage/\pageref{LastPage}}%当前页，记得调用前文提到的宏包
\lfoot{Zhejiang University}
\rfoot{School of Mathematical Sciences}
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    colorlinks=true,%链接将会有颜色，默认是红色
    linkcolor=blue,%内部链接，那些由交叉引用生成的链接将会变为蓝色（blue）
    filecolor=magenta,%链接到本地文件的链接将会变为洋红色（magenta）
    urlcolor=blue,%链接到网站的链接将会变为蓝绿色（cyan）
    }

\newtheorem{theorem}{Theorem}
\newtheorem{proof}{Proof}
\newtheorem{solution}{Solution:}

\title{\textbf{Find the critical node}}
\date{\today}

\begin{document}
\subsection*{\uppercase\expandafter{\romannumeral 1} Find the critical node}
$$\begin{pmatrix}
    Y_{LL} & Y_{LG} \\
    Y_{GL} & Y_{GG}
  \end{pmatrix}\begin{pmatrix}
                 V_L \\
                 V_G
               \end{pmatrix} = \begin{pmatrix}
                                 I_L \\
                                 I_G
                               \end{pmatrix}$$
Then 
$$Y^{-1}_{LL}\begin{pmatrix}
    Y_{LL }& Y_{LG}
  \end{pmatrix}\begin{pmatrix}
                 V_L \\
                 V_G
               \end{pmatrix} = Y_{LL}^{-1}T_L\Rightarrow V_L + Y_{LL}^{-1}Y_{LG}V_G = Y^{-1}_{LL}I_L$$
Take the $k_{th}$ row, multiply $V^*_k$ and divide $(Y_{LL}^{-1})_{kk}$, we have
$$\begin{aligned}
  &\left(\frac{1}{(Y_{LL}^{-1})_{kk}}V_k+\sum_{j=1}^{m}\frac{(Y_{LL}^{-1}Y_{LG})_{kj}}{(Y_{LL}^{-1})_{kk}}V_{n+j}\right)V^*_k\\
&= \sum_{j=1}^{m}\frac{(Y_{LL}^{-1})_{kj}}{(Y_{LL}^{-1})_{kk}}\frac{V^*_k}{V^*_j}I_jV^*_j\\
&= \sum_{j=1}^{m}\frac{(Y_{LL}^{-1})_{kj}}{(Y_{LL}^{-1})_{kk}}\frac{V^*_k}{V^*_j}S^*_j
\end{aligned}$$

Take $k = 1$, 
\begin{equation}(V_L)_1+\sum_{j=1}^{m}(Y_{LL}^{-1}Y_{LG})_{1j}(V_G)_j= \sum_{j=1}^{m}(Y_{LL}^{-1})_{1j}I_j\end{equation}
That is
$$\left(\frac{1}{(Y_{LL}^{-1})_{11}}(V_L)_1+\sum_{j=1}^{m}\frac{(Y_{LL}^{-1}Y_{LG})_{1j}}{(Y_{LL}^{-1})_{11}}(V_G)_j\right)(V_L)^*_1
= \sum_{j=1}^{m}\frac{(Y_{LL}^{-1})_{1j}}{(Y_{LL}^{-1})_{11}}I_j(V_L)_1^*$$\\

Denote $S^*_c = \sum_{j=1}^{m}\frac{(Y_{LL}^{-1})_{1j}}{(Y_{LL}^{-1})_{11}}I_j(V_L)_1^*,\psi_1 = \frac{1}{(Y_{LL}^{-1})_{11}}, \varphi_j = \frac{(Y_{LL}^{-1}Y_{LG})_{1j}}{(Y_{LL}^{-1})_{11}}$\\

Then we have 
\begin{equation}\psi_1\left|(V_L)_1\right|^2 + \sum_{j=1}^{m}\varphi_j(V_G)_j(V_L)_1^* = S^*_c\end{equation}
Take conjugate transpose, then
\begin{equation}
  \psi_1^*\left|(V_L)_1\right|^2 + \sum_{j=1}^{m}\varphi_j^*(V_G)^*_j(V_L)_1 = S_c
\end{equation}
Multiply (2) and (3), divided by $|\psi_1|^2$, we have
\begin{equation}\left|(V_L)_1\right|^4+\left(2Re\left(\frac{\sum_{j=1}^{m}\varphi_j(V_G)_j \psi_1^*(V_L)^*_1}{|\psi_1|^2}\right)+\left|\sum_{j=1}^{m}\frac{\varphi_{1j}}{|\psi_1|}(V_G)_j\right|^2\right)\left|(V_L)_1\right|^2
  -\left(\frac{|S_c|}{|\psi_1|}\right)^2=0\end{equation}

Since
\begin{equation}\begin{aligned}&\frac{\sum_{j=1}^{m}\varphi_j(V_G)_j \psi_1^*(V_L)^*_1}{|\psi_1|^2}\\
  =&\frac{\sum_{j=1}^{m}\varphi_j(V_G)_j(V_L)^*_1}{\psi_1}\\
  =&\sum_{j=1}^{m}\frac{(Y_{LL}^{-1}Y_{LG})_{1j}}{(Y_{LL}^{-1})_{11}}(V_G)_j(V_L)^*_1(Y_{LL}^{-1})_{11}\\
  =& \left(\sum_{j=1}^{m}(Y_{LL}^{-1}Y_{LG})_{1j}(V_G)_j\right)(V_L)^*_1
\end{aligned}
\end{equation}
and by (1), we have 
\begin{equation}
  \begin{aligned}
    (5) &= \left(\sum_{j=1}^{m}(Y_{LL}^{-1})_{1j}I_j-(V_L)_1\right)(V_L)^*_1\\
    &= \sum_{j=1}^{m}(Y_{LL}^{-1})_{1j}I_j(V_L)^*_1-\left|(V_L)_1\right|^2\\
    &= \frac{S^*_c}{\psi_1}-\left|(V_L)_1\right|^2
  \end{aligned}
\end{equation}

Apply to (4), then we have
\begin{equation}
  \left|(V_L)_1\right|^4-\left(2Re\left(\frac{S^*_c}{\psi_1}\right)+\left|\sum_{j=1}^{m}\frac{\varphi_{1j}}{|\psi_1|}(V_G)_j\right|^2\right)\left|(V_L)_1\right|^2
  +\left(\frac{|S_c|}{|\psi_1|}\right)^2=0
\end{equation}

Which means (7) has positive solutions for $\left|(V_L)_1\right|$

Take $\Phi = \left|\sum_{j=1}^{m}\frac{\varphi_{1j}}{|\psi_1|}(V_G)_j\right|^2$ and $S_c = S_0 +\lambda S_1$, hence 
$$\begin{aligned}
  \Delta &= \left(2Re\left(\frac{S^*_c}{\psi_1}\right)+\Phi\right)^2 - 4\left(\frac{|S_c|}{|\psi_1|}\right)^2= 4Re\left(\frac{S^*_c}{\psi_1}\right)\Phi+\Phi^2
  -4Im^2\left(\frac{S^*_c}{\psi_1}\right)\\
  &=-4Im^2\left(\frac{S^*_1}{\psi_1}\right)\lambda^2+\left(-8Im\left(\frac{S^*_0}{\psi_1}\right)Im\left(\frac{S^*_1}{\psi_1}\right)
  +4Re\left(\frac{S^*_1}{\psi_1}\right)\Phi\right)\lambda+\Phi^2+4Re\left(\frac{S^*_0}{\psi_1}\right)\Phi-4Im\left(\frac{S^*_c}{\psi_0}\right)\geq 0
\end{aligned}$$

Hence, $$\lambda \in \left(\frac{-B-\sqrt{B^2-4AC}}{2A},\frac{-B+\sqrt{B^2-4AC}}{2A}\right)$$
where $$\left\{\begin{aligned}
  A & = -4Im^2\left(\frac{S^*_1}{\psi_1}\right)\\
  B &= -8Im\left(\frac{S^*_0}{\psi_1}\right)Im\left(\frac{S^*_1}{\psi_1}\right)+4Re\left(\frac{S^*_1}{\psi_1}\right)\Phi\\
  C &= \Phi^2+4Re\left(\frac{S^*_0}{\psi_1}\right)\Phi-4Im\left(\frac{S^*_0}{\psi_0}\right)
\end{aligned}\right.$$
\end{document}